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1
Chapter 1 : MATTER
1.1 Atoms and Molecules
1.2 Mole Concept
1.3 Stoichiometry
Chapter 1 : MATTER
1.1 Atoms and Molecules
1.2 Mole Concept
1.3 Stoichiometry
2
Learning Outcomes
(a) Describe proton, electron and neutron in terms of mass and charge
(b) Define proton no, nucleon no and isotope
(c) Write isotopic notation
(d) Define relative atomic mass and relative molecular mass
(e) Calculate average atomic mass of an element
3
At the end of the lesson, you should be ABLE to:
MATTER
Solutions
Mixtures
4
5
What is ‘matter’?
Anything that occupies space/has volume and
has mass.
Example:
………………………………………………………..
Matter consists of 3 types of particles: atoms, molecules or ions.
Give an example of non-matter. Give your reason.
3 Types of Particles
ATOMSAn atom is the
smallest unit of a
chemical
element/compound.
Examples:
Na, Cu, Ne
MOLECULESA molecule consist of a
small number of atoms
joined together by bonds
Examples:
O2, CO2, H2O
IONSAn atom or molecule
with a net electric
charge due to the
loss or gain of one or
more electrons.
Examples:
Na+, NH4+,
O2-, NO3-
Example: Cuatom
6
7
Atoms
8
Diatomic Molecules
(e)
(n)
(p)
Atomic Structure
An atom consist of 3
subatomic particles:
Proton (p)
Neutron (n)
Electron (e) 9
Subatomic Particles
Subatomic
particle
Mass
(gram)
Charge
(coulomb)
Relative
charge
Electron
(e)9.1 x 10-28
-1.6 x 10-19
Proton
(p)1.67 x 10-24
+1.6 x 10-19
Neutron
(n)1.67 x 10-24 0
10
• Proton number (Z) :
………………………………………………………..
• Nucleon number (A) :
…………………………………………………………
Proton number (Z) & Nucleon number (A)
11
12
Element
symbol
(can be an
atom or an
ion!!!)
Proton
number
(p only)
Nucleon
number
(p + n)
An atom can be represented by an isotope
notation (atomic symbol) :
Isotope Notation
13
Total charge
on the ion
proton number of
mercury,
Z = 80
Nucleon number of
mercury, A = 202
The number of
neutrons
= A – Z
= 202 – 80
= 122
* No of e = 80 -2 = 78
14
Isotopes
..............................................................................................
..............................................................................................
............................................................................................
• Examples:
Hg200
80
2
1H U235
92
U238
92 3
1H
Hg202
80
15
Ions (cation & anion)
Atom
(neutral)
Cation
(+ ion)
Anion
(- ion)Atom
loses e
Atom
accepts e
NaNa+
+ e Cl + e Cl-
Z = 11e = 11
Z = 11e = 10
Z = 17e = 17
Z = 17e = 18
16
Exercise 1
Symbol Number of :
ChargeProton Neutron Electron
• Give the number of protons, neutrons, electrons and charge in each of the following species:
Cu63
29
217
8O
359
27Co
Hg200
80
17
Exercise 2
Species
Number of :Notation
for nuclideProton Neutron Electron
A 2 2 2
B 1 2 0
C 1 1 1
D 7 7 10
• Write the appropriate notation for each of the following nuclide :
18
Learning Outcomes
(a) Describe proton, electron and neutron in terms of mass and charge
(b) Define proton no, nucleon no and isotope
(c) Write isotopic notation
(d) Define relative atomic mass and relative molecular mass
(e) Calculate average atomic mass of an element
19
At the end of the lesson, you should be ABLE to:
√√√
NEXT….
Relative Mass
Relative Atomic Mass (Ar)
Relative Molecular Mass (Mr)
Average mass of one
atom of an element
compared to 1/12
mass of one 12C atom
with the mass 12 g.
A mass of one molecule
of a compound compared
to 1/12 mass of one
atom
of 12C with the mass
12 g.
Ar =
Mr =
Both Ar and Mr are dimensionless; have no unit.
Relative Mass
Relative Atomic Mass (Ar)
Relative Molecular Mass (Mr)
Amu =atomic mass unit
Example 1
12-C of atom one of Mass 12
1
Y of atom one of mass Average
Determine the relative atomic mass of an element Y if the
ratio of the atomic mass of Y to carbon-12 atom is 0.45.
ANSWER :
Ar =
= 12 x 0.45
1
= 5.4
23
The relative molecular mass of a compound is the SUMMATION of
the relative atomic masses of all atoms in a molecular formula.
NaOHAr of Na = 23.0
Ar of O = 16.0
Ar of H = 1.0
EXAMPLE 2:
Mr of NaOH= 23.0 + 16.0 + 1.0= 40
24
Example 3 :
Calculate the relative molecular mass (Mr) of C5H5N
Answer :
Mr =
Learning Outcomes
(a) Describe proton, electron and neutron in terms of mass and charge
(b) Define proton no, nucleon no and isotope
(c) Write isotopic notation
(d) Define relative atomic mass and relative molecular mass
(e) Calculate average atomic mass of an element
25
At the end of the lesson, you should be ABLE to:
√√√
NEXT….
√
Device used to measure Ar : Mass Spectrometer
Device used to measure Ar : Mass Spectrometer
Mass Spectrum
Mass Spectrum of Magnesium
• The mass spectrum of Mg shows that Mg consists of 3 isotopes : 24Mg, 25Mg and 26Mg.
• The height of each line is proportional to the abundance of each isotope.
• 24Mg is the most abundant of the three isotopes.
63
8.19.1
24 25 26
Rela
tiv
e a
bu
nd
an
ce
(Q
)
m/e
(mass, amu)
% or
Ratio or
Fraction
m/e =
mass/charge
How to calculate the average atomic mass from mass spectrum?
Q = the relative / fractional / ratio / percentage abundance of isotopes of the element in the mixture
m = the isotopic mass of the element in unit amu.
The unit of average atomic mass is amu
Amu =atomic mass unit
Average
atomic
mass=
Example 1
Copper, Cu consists of two isotopes which is 69%
63Cu and 31% 65Cu. The isotopic mass of 63Cu and
65Cu are 62.9 a.m.u and 64.9 a.m.u respectively.
Calculate the average atomic mass of copper.
Average
atomic mass=
∑ Qi mi
∑ Qi
= (69 x 62.9 a.m.u) + (31 x 64.9 a.m.u)
69 + 31
= 63.52 a.m.u
69
31
62.9 64.9
Rela
tive
ab
un
dan
ce
(%)m/e (amu)
SOLUTION:
Example 2
a. What isotopes are present in Rb?
b. What is the percentage abundance of each isotope?
c. Calculate the relative atomic mass (Ar) of Rb.
18
7
85 87
Re
lati
ve
ab
un
da
nc
e
m/e
(amu)
Given below is a mass spectrum of rubidium element, Rb.
33
SOLUTION
18
7
85 87
Re
lati
ve
ab
un
da
nc
e
m/e
(amu)
(a) 85Rb and 87Rb
(b) % abundance 85Rb
= 18 X 100 = 72%
(18+7)
% abundance 85Rb
= 7 X 100 = 28%
(18+7)
85.56
amu x12.0012
1
amu 85.56Rb of A
amu 85.56
25
)87x7()85x18(
Qi Rb of mass Average
r
iiMQ
CONTINUE SOLUTION
C.
The ratio of relative abundance of naturally occurring of chlorine
isotopes is as follow :
Based on the carbon-12 scale, the relative isotopic mass of
35Cl = 34.9689 and 37Cl = 36.9659. Calculate the Ar of
chlorine.
Example 3
3.127 37
35
Cl
Cl
SOLUTION
CONTINUE SOLUTION
Example 4
The relative atomic mass of Li and Li
are 6.01 and 7.02 respectively.
What is the percentage abundance of each
isotope if the relative atomic mass of Li is
6.94?
63
73
SOLUTION
CONTINUE SOLUTION
41
RecapWhat have you learnt today?